Question regarding a critical point formula Hey everyone, I fear that on my final I'll be given an equation that ressembles , and I won't be able to find the critical points. I know that when you have the form , you can use . Is there such a method for ? 
Re: Question regarding a critical point formula There is a cubic formula http://en.m.wikipedia.org/wiki/Cubic_function But it is a bit complicated to use so even if you get a cubic function it is better to use the techniques of precalculus to determine rational roots (if any). :D 
Re: Question regarding a critical point formula ouch, I don't think I'll be able to memorize that formula in time haha. Alright, I'll use other methods. Thanks agentredlum! 
Re: Question regarding a critical point formula "Solution: Given ax³+bx²+cx+d let f(x)= ax³+bx²+cx+d f'(x)=3ax²+2bx+c,where f'(x)=0 then there are two critical point ,and at least two real roots there for we can use the formula x=[2b±v(4b²12ac)]/6a" 
Re: Question regarding a critical point formula Quote:

Re: Question regarding a critical point formula Quote:
IF the original function were a cubic, then the derivative would be a quadratic but the OP ways he/she solved a quadratic before. I would interpret this as asking what to do if the derivative itself is a cubic polynomial. MFP, I doubt that you would be expected to know the cubic formula. IF the derivative turns out to be cubic, try simple integer values for x bet on it being easily factorable. 
Let c = p  q. Assume a ? 0, b, p and q are real numbers. However if we have ax² + bx + p = q then we have ax² + bx + p  q = 0. The typical quadratic formula would be . To prove the quadratic formula is quite easy e.g. divide both sides by a then . . . The quadratic formula is true if and only if ax² + bx + c = 0. 
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